Question

if the first term of an ab is -5 and the common difference is 2 then the sum of​

Answer 1

Answer:

$sn = \frac{n}{2} (2a + (n - 1)d)$

$sn = \frac{n}{2} ( - 10 + (n - 1)2)$

$sn = \frac{n}{2} ( - 10 + 2n - 2)$

$sn = \frac{n}{2} ( 2n - 12)$

$sn = \frac{ {2n}^{2} - 12n}{2}$

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Answer 2

Answer:

11

Step-by-step explanation:

a=11

d=2

let number of terms be n+1

an+an-1=60

a+(n-1) d+a+ (n+1-1) d=60

11+(n-1) 2+11+n (2) = 60

22+2n-2+2n=60

4n=60-20=40

n=40/4=10

Number of terms are 11